Question: Factor the following expression: $-4$ $x^2+$ $1$ $x+$ $3$
This expression is in the form ${A}x^2 + {B}x + {C}$ . You can factor it by grouping. First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-4)}{(3)} &=& -12 \\ {a} + {b} &=& & & {1} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-12$ and add them together. Remember, since $-12$ is negative, one of the factors must be negative. The factors that add up to ${1}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-3}$ and ${b}$ is ${4}$ $ \begin{eqnarray} {ab} &=& ({-3})({4}) &=& -12 \\ {a} + {b} &=& {-3} + {4} &=& 1 \end{eqnarray} $ Next, rewrite the expression as ${A}x^2 + {a}x + {b}x + {C}$ $ {-4}x^2 {-3}x +{4}x +{3} $ Group the terms so that there is a common factor in each group: $ ({-4}x^2 {-3}x) + ({4}x +{3}) $ Factor out the common factors: $ x(-4x - 3) - 1(-4x - 3) $ Notice how $(-4x - 3)$ has become a common factor. Factor this out to find the answer. $(-4x - 3)(x - 1)$